A332878 a(1)=1; a(n+1) is the smallest number not yet seen in the sequence which is divisible by 1 + Omega(a(n)).

1, 2, 4, 3, 6, 9, 12, 8, 16, 5, 10, 15, 18, 20, 24, 25, 21, 27, 28, 32, 30, 36, 35, 33, 39, 42, 40, 45, 44, 48, 54, 50, 52, 56, 55, 51, 57, 60, 65, 63, 64, 7, 14, 66, 68, 72, 78, 76, 80, 84, 70, 88, 75, 92, 96, 49, 69, 81, 85, 87, 90, 95, 93, 99, 100, 105, 104

Offset: 1

Author

David James Sycamore, Feb 28 2020

Keywords

• nonn

Comments

Here Omega = A001222, the number of prime factors counted with multiplicity.

Conjecture: This is a permutation of the natural numbers in which the primes appear in their natural order. Prime p > 2 arises as a(k) if and only if a(k-1) = 2^(p-1), in which case a(k+1) = 2*p. The sequence of numbers k such that a(k) is prime starts 2, 4, 10, 42, ... How does it continue?

a(636) = 11, a(2530) = 13, a(39731) = 17. It appears that the prime p occurs roughly at index 2^(p-2)*(1 + O(1/log p)). It is followed by 2p and then a multiple of 3. The graph of the sequence has several "branches" which can be labeled by odd primes: Most numbers occur on the main (p=3) branch which has an initial slope of about 1.61 increasing to 1.65 in the range 1e4 .. 4e4. A smaller fraction of the numbers lie on a second (p=5) and third (p=7) branch with slope of roughly 1.25 resp. 1.11 around n ~ 4e4, and a very small fraction lies on the branches with even lower slope (about 0.15 for the p=11 and 0.035 for the p=13 branch). - M. F. Hasler, Mar 04 2020

Examples

			a(1) = 1 => 1 + Omega(a(1)) = 1, so a(2) must be 2.
Then 1 + Omega(a(2)) = 2, so a(3) = 4.
Since 1 + Omega(4) = 3, a(4) = 3; then 1 + Omega(3) = 2 and a(5) = 6, etc.
		

Links

• M. F. Hasler, Table of n, a(n) for n = 1..10000
• M. F. Hasler, Scatterplot of a(1..40000).

Crossrefs

Cf. A000040, A001222, A061286.

Programs

Mathematica

a[1]=1; a[n_] := a[n] = Block[{s = Array[a, n-1], k, o = 1 + PrimeOmega@ a[n-1]}, k = o; While[ MemberQ[s, k], k += o]; k]; Array[a, 56] (* Giovanni Resta, Mar 04 2020 *)

PARI

{A332878_vec(N,a=1,u=1)=vector(N,n, a=n=bigomega(a)+1; while(bittest(u,a), a+=n); u+=1<M. F. Hasler, Mar 04 2020